Problem: N different natural numbers are given. If you select four different ones and set them as $ A $, $ B $, $ C $, $ D $, the following formula

$ \ Frac {A + B} {C --D} $



I want to find the maximum value of.




Given N different natural numbers, choose 4 different from them and create a program to find the maximum value of the above formula.



Input

The input is given in the following format.


N
a1 a2 ... aN


The number N (4 ≤ N ≤ 1000) of natural numbers is given in the first line. The value ai (1 ≤ ai ≤ 108) of each natural number is given in the second line. However, the same natural number does not appear more than once (ai ≠ aj for i ≠ j).

Output

Outputs the maximum value of the above formula as a real number for a given N natural numbers. However, the error must not exceed plus or minus 10-5.

Examples

Input

10
1 2 3 4 5 6 7 8 9 10


Output

19.00000


Input

5
22 100 42 3 86


Output

9.78947


Input

6
15 21 36 10 34 5


Output

18.00000


Input

4
100000 99999 8 1


Output

28571.285714

Solution (in CPP):

Solution:
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
typedef long long int ll;

using namespace std;

struct Info{

	Info(double arg_left,double arg_right,double arg_div){
		left = arg_left;
		right = arg_right;
		div = arg_div;
	}

	bool operator<(const struct Info &arg) const{
		return div > arg.div;
	};

	double left,right,div;
};

int main(){

	int N;
	double top4[4] = {0,0,0,0};
	scanf("%d",&N);

	double table[N];

	for(int i = 0; i < N; i++)scanf("%lf",&table[i]);

	sort(table,table+N);

	top4[0] = table[N-1];
	top4[1] = table[N-2];
	top4[2] = table[N-3];
	top4[3] = table[N-4];

	priority_queue<Info> Q;

	for(int i = 1; i < N; i++){
		Q.push(Info(table[i],table[i-1],table[i]-table[i-1]));
	}

	int count = 0;
	double maximum = 0;

	while(Q.empty() == false && count <= 50){
		for(int i = 0; i < 3; i++){
			for(int k = i+1; k < 4;k++){
				if(Q.top().left != top4[i] && Q.top().left != top4[k] &&
						Q.top().right != top4[i] && Q.top().right != top4[k]){
					maximum = max(maximum,(top4[i]+top4[k])/Q.top().div);
				}

			}
		}
		Q.pop();
	}

	printf("%.5lf\n",maximum);

    return 0;
}